WAEC 2011 · Paper 2 · Q1

(a)
Simplify: $\dfrac{\frac12 \text{ of } \frac14 \div \frac13}{\frac16 - \frac34 + \frac12}$ .
(b)
Given that $\sqrt{x} = 10^{\bar1.6741}$ , without using calculators, find the value of $x$ .

Worked solution (try it first)

(a)

“Of” means multiply, and it comes before the division: $\frac12 \text{ of } \frac14 = \frac18$ .
Dividing by $\frac13$ is multiplying by 3: $\frac18 \times 3 = \frac38$ .
This is the numerator.
Denominator over the LCM 12: $\frac{2}{12} - \frac{9}{12} + \frac{6}{12} = -\frac{1}{12}$ .
Divide: $\frac38 \div \left(-\frac{1}{12}\right) = \frac38 \times (-12)$
$= -\frac92$ .
So the value is $-4\frac12$ .

(b)

Take logs of both sides: $\log\sqrt{x} = \bar1.6741$ , that is $\frac12\log x = \bar1.6741$ .
Multiply by 2, doubling the characteristic and the mantissa separately: $2 \times (-1) = -2$ and $2 \times 0.6741 = 1.3482$ .
So $\log x = -2 + 1.3482 = \bar1.3482$ .
Read the antilog of $\bar1.3482$ from the tables: $x = 0.2229$ .

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